SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Since $q=a_1^2+b_1^2$ then $q$ must have the form $4k+1$; otherwise, $a_1^2+b_1^2$ is divisible by $q=4k+3$ which implies that $q\mid a_1,q\mid b_1$, a contradiction.

Note that $q=a_2^2+2b_2^2$ so in modulo $8$, number $q$ is congruent to either $1$ or $3$ modulo $8$, but $q\equiv 1(\mathrm{mod}4)$ then $q\equiv 1(\mathrm{mod}8)$.

Continue, $q=a_3^2+3b_3^2$ so by taking modulo $3$, it gives $q$ is congruent to $1$ in modulo $3$. Similarly, by considering two equations $q=a_5^2+5b_5^2$ and $q=a_7^2+7b_7^2$, we have $q\equiv 1,4(\mathrm{mod}5)$ and $q\equiv 1,2,4(\mathrm{mod}7)$.

These congruences can be put together as $q$ is congruent to $$1,121,169,289,361,529\text{ in modulo }840.$$ The first smallest values that satisfy the previous congruence are $$1,121,361,169,289,529,841,961,1009,\dots$$ and it is easy to check that the smallest prime solution is $1009$.

And it does satisfy the problem requirement by the following equalities $$\begin{array}{rl}& 1009\\ & =28^2+15^2=19^2+2(18^2)\\ & =31^2+3(4^2)=15^2+4(19^2)\\ & =17^2+5(12^2)=25^2+6(8^2)\\ & =1^2+7(12^2)=19^2+8(9^2)\\ & =28^2+9(5^2)=3^2+10(10^2)\end{array}$$ Therefore, the answer is $1009$.